Definition of the nth derivative? [First post]

Question

If the definition of the derivative is $$ f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} $$ Would it make sense that the nth derivative would be (I know that the 'n' in delta x to the nth power is useless) $$ f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n} $$ I came to this conclusion using this method $$ f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} $$ (this is correct right?) $$ f^{\prime\prime}(x) = \lim_{\Delta x \to 0} \dfrac{f^\prime(x+\Delta x) - f^\prime(x)}{\Delta x}=$$
$$\lim_{\Delta x \to 0}\dfrac{\dfrac{f((x+\Delta x)+\Delta x)-f(x+\Delta x)}{\Delta x}-\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x}=$$
$$\lim_{\Delta x \to 0}\dfrac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2} $$ After following this method a couple of times(I think I used it to the 5th derivative) I noticed the pattern of $$(a-b)^n$$ And that is how i arrived at $$ f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n} $$ Have I made a fatal error somewhere or does this definition actually follow through?
Thanks for your time I really appreciate it.
P.S. Any input on using tags will be appreciated.